JMC 2018

Halves them to 45°

HintA square's diagonal is a line of symmetry.

The whyBecause the diagonal is a line of symmetry, it bisects the right angles at both ends, giving 45° each way and making two right-angled isosceles triangles. Angles at a corner shared by a square and another shape are then just a matter of adding.

7; 8; 10; 12

HintGreek number words: hepta, octo, deka, dodeka.

The whyRegular polygon questions almost always need the number of sides for a perimeter or an angle. A regular polygon has equal sides, so its perimeter is the side length times the number of sides.

b − a + 1

HintCount from 3 to 5 on your fingers before you subtract.

The whyFrom 12 to 30 inclusive there are 30 − 12 + 1 = 19 numbers, not 18. Subtracting alone counts the gaps; the +1 adds back the number you started on. Strictly between a and b there are b − a − 1. The same fencepost logic counts terms in a sequence and posts along a fence.

8

HintScale two thirds up until its denominator is twelve.

The whyTo place a fraction between two others, rewrite them all over one common denominator: 1/2 = 6/12 and 2/3 = 8/12, so 7/12 sits between them. With the same numerator instead, the fraction with the larger denominator is the smaller one — one eighth is less than one quarter because the whole is cut into more pieces.

6

HintCount the faces of a dice.

The whyEach face is a × a and there are six of them. For a solid with no hidden pockets, count the faces you can see from each of the six directions — top, bottom and four sides — rather than the faces of each cube.

5

HintOne test looks at the last digit.

The whyBecause 3 and 5 have no common factor, being a multiple of both is the same as being a multiple of 15. So check the last digit (0 or 5) and the digit sum (a multiple of 3). The same works for 45 = 9 × 5 and 12 = 3 × 4, but not for 12 = 2 × 6, because 2 and 6 share a factor.

The leading digit

HintThe most significant place decides the most.

The whyFix the front of the number first, then fill greedily: for the largest, the biggest remaining digits in decreasing order; for the smallest, the smallest remaining in increasing order. For 'closest together', give the two numbers neighbouring front digits, then push one down and the other up.

11

HintSix squares joined edge to edge that fold up with no overlap — there are more than you'd guess.

The whySix of the eleven have a strip of four squares with one square on each side. To test a net quickly, check that no four squares meet at a point (a 2 × 2 block) and that the two 'flap' squares fold onto opposite faces, not the same one.

Three rectangles and two triangles

HintPicture a tent-shaped box unfolded flat: the long panels, plus the piece at each end.

The whyA net must contain every face exactly once and fold up with no overlap. For a cube that means six squares; for a triangular prism, three rectangles and two triangles; for a square-based pyramid, one square and four triangles.

4

HintThink about the worst case: how many picks can go by with no match at all?

The whyOne more than the number of colours forces a pair, because the worst case is one of each colour before the next pick must repeat one. Always reason from the worst case: what is the most you could pick without success?

360° ÷ n

HintAll the exterior angles are equal and together make one full turn.

The whyFor a regular polygon, exterior angle = 360° ÷ n and interior angle = 180° − exterior. Running it backwards is how you find n: an interior angle of 156° means an exterior angle of 24°, so n = 360 ÷ 24 = 15.

All sides equal and all angles equal

HintThe one word that lets you assume everything is identical.

The why'Regular' is the word that unlocks a diagram: it hands you equal sides (so isosceles triangles from the centre) and equal angles (so 360° ÷ n exterior, and 180° minus that interior). Without it, none of those follow.

b÷c

HintDivide each part of the sum by c separately, then add the results.

The whyBecause division distributes over addition, (a + b) ÷ c splits into a÷c + b÷c — often easier mental arithmetic than adding first. This only works for dividing a SUM by a number: a ÷ (b + c) does NOT equal a÷b + a÷c, so always check which value is being divided.

Combine the losses and gains into one net change first

HintAdd the joiners and subtract the leavers, in either order.

The whyWork out the net change as one signed number (arrivals minus departures) and apply it once, rather than updating a running total after every single step. If the net change is negative, subtract it from the starting count; if positive, add it. One combined calculation beats several small ones.

Convert the full hours to minutes (×60), then add the leftover minutes

HintBreak the gap into a round number of something, plus what's left.

The whySplit the interval into a whole number of hours plus a remaining number of minutes, then convert: total minutes = hours × 60 + remaining minutes. It's easy to slip when the times sit either side of an hour mark, so sanity-check your hour count against a clock face if you're unsure.

Have equal totals, so their other entries can be compared directly

HintEvery row, column and diagonal in the square adds to the same fixed amount.

The whyEvery row, column and diagonal in a magic square shares one fixed total. When two lines cross the same cell, setting their sums equal and cancelling the shared entry links the OTHER numbers on those lines directly — often solving the puzzle without ever finding the magic total itself.

Estimate using powers of ten by rounding the numbers first

HintThink in tens, hundreds and thousands rather than exact figures.

The whyWhen a question asks for an approximate size or how many digits an answer has, round the numbers involved to the nearest power of ten (or a friendly multiple) before dividing or multiplying — you rarely need the exact value. For a division, comparing digit counts (or trailing zeros) is often the fastest route to the digit-count of the answer.

Add up all the numbers, then subtract the given total

HintSum everything first — the leftover gap is your answer.

The whyAdding the whole set (including the odd one out) is usually easier than trialling each candidate in turn. Subtract the known target total from that full sum, and whatever's left over is the value that was excluded.

Let them be 2k and 3k, then solve k from the angle sum

HintTurn the ratio into algebra with one shared unknown multiplier.

The whyWhenever a diagram splits an angle (or several angles) in a given ratio, introduce a single unknown k so the parts become multiples of k — 2k and 3k for a 2:3 split. Then use whichever angle-sum fact applies (180° in a triangle, 360° in a quadrilateral, or two triangles' sums linked by a given relationship) to write one equation and solve for k.

Combine the multiples into one equation, then find the integer solution

HintOnly whole-number pairs where every count stays positive will work.

The whyGive each unknown its own letter (the 'unit' amounts), express every quantity as a multiple of one of them, and add them to the given total to get one equation in two unknowns. Because every quantity must be a positive whole number, only a small number of pairs actually satisfy it — checking them is faster than solving generally.

cells per piece

HintThink about how many identical shapes could ever squeeze into the whole grid.

The whySince each piece covers a fixed number of cells and pieces can't overlap, the count can never exceed total cells ÷ cells per piece, rounded down. That's only an upper bound — you still need an arrangement that actually achieves it — but it instantly rules out any higher answer.

Gives the length of the other side

HintTwo numbers multiply together to make the space a shape covers.

The whyA rectangle's area equals length × width, so any one side equals area ÷ the other side. When two rectangles in a diagram share an edge, find that shared length from the rectangle whose area you already know, then use it to solve the second rectangle.

self-referential

HintIt talks about its own answer, not something separate.

The whyA self-referential statement's truth depends on a fact about itself — like a sentence claiming a count that only holds for the right number. You can't calculate the answer directly; instead you test candidates and see which one makes the statement true of itself.

Rearrange the sum rule to isolate the missing earlier term

HintMove every known value to one side, leaving the gap alone.

The whyIf term n equals the sum of the k terms before it, rearrange that rule to express any one of those earlier terms as (term n) minus the sum of the others. This works backwards through a sequence just as easily as forwards, so 'find the earlier blank' is really rearrange-and-subtract, not a fresh calculation from scratch.

two matching edges; smaller than the sum of the separate perimeters

HintWherever two rectangles touch, that boundary disappears from what you can see around the outside.

The whyEvery rectangle contributes its own full perimeter on its own — but wherever two are joined, the touching edges sit inside the shape rather than around its outline, so that length is removed from both rectangles at once. Count how many joins there are, work out what each one removes, and subtract that total from the sum of the individual perimeters.